Question

Va rog urgent! Mulțumesc!❤❤

Answer 1

a)
1+2+3+...+n=n(n+1)/2
Pt. k=1 ==> 1=1(1+1)/2/ 1=1 => P(1) A
Ec. 1+2+3+...+k=k(k+1)/2
Pt. k=k+1 Ec. devine ==> 1+2+3+...+k+(k+1)= (k+1)(k+2)/2
k(k+1)/2+ k+1 = (k+1)(k+2)/2
[ k(k+1) +2(k+1) ]/2 =(k+1)(k+2)/2
(k+1)(k+2)/2 = (k+1)(k+2)/2  A,∀ k ∈ N* => A,n ∈ N*
b)
Pt. k=1 => 4-3 = 1(2-1) - 1=1 => P(1) A
Ec. 1+5+9+   +(4k-3)=k(2k-1)
Pt. k=k+1 Ec. devine==> 1+5+9+  +(4k-3)+(4k+1)=(k+1)(2k+1)
k(2k-1)+4k+1=(k+1)(2k+1)
2k²-k+4k+1=2k²+k+2k+1
2k²+3k+1=2k²+3k+1  A,∀ k ∈ N* => A,n ∈ N*
c)
1²+2²+3²+  +n²= n(n+1)(2n+1)/6
Pt. k=1 => 1²=1(1+1)(2+1)/6 1=6/6 1=1 ==>P(1) A
Ec. 1²+2²+3²+ +k²=k(k+1)(2k+1)/6
Pt. k=k+1 Ec. devine==> 1²+2²+3²+ +k²+(k+1)²=(k+1)(k+2)(2k+3)/6
k(k+1)(2k+1)/6 +(k+1)²= (k+1)(k+2)(2k+3)/6
k(k+1)(2k+1)+6(k+1)² /6 =(k+1)(k+2)(2k+3)/6
(k+1)(2k²+k+6k+6) /6 =(k+1)(k+2)(2k+3)/6
(k+1)(k+2)(2k+3)/6=(k+1)(k+2)(2k+3)/6  A,∀ k ∈ N* => A,n ∈ N*